User blog:Cheetahrock63/Uniform tilings with linear pseudogonal faces
(incomplete) Pseudogons are an interesting bunch with many wonders such as the 2i -gon, the 3i -gon and the 4i -gon— they're the "imaginarygons". They're apeirogons that are circumscribed by a hypercycle rather than a horocycle (traditional apeirogon \{\infty\} ) or a circle (failed star polygons such as the \sqrt{2} -gon, \pi -gon, and e -gon— the irrationalgons). Informally explained, you can construct the set of vertices of a pseudogon, an \frac{i \pi}{\lambda} -gon, by considering a hyperbolic straight line L , partitioning the line into segments of length 2 \lambda , translating all the vertices of the line segments some distance d away from the straight line all along a line perpendicular to L . The boundary is formed by joining every vertex with the two closest vertices by line segments. The interior is defined to be the set of all points on one side of the hyperbolic plane defined by the boundary or the other. I suppose that “'hyperbolic half-plane'” isn’t too inappropriate of a name for the shape since a line divides the plane into two of them, but you can fit up to a good infinite number of “half-planes” in this plane with room to spare so the name might be a little bit misleading. You may notice that the set that consists of two consecutive vertices on L and their translated cousins form the vertices of a Saccheri quadrilateral, a quadrilateral in the hyperbolic plane with two right angles, with a base length of 2\lambda and d as the edge lengths of the two sides perpendicular to the base. Bisect that and you've got two Lambert quadrilaterals with their shortest edges having length \lambda and their longest edges having length d . If you know how to solve the quadrilateral(s), you can figure out the edge length, internal angle measure, and their relationship of your constructed pseudogon. Consider the relationship between a p -gon's edge length s and internal angle \theta in the hyperbolic plane, s = 2 {\rm {arccosh}} (|\frac{\cos (\frac{\pi}{p})}{\sin (\frac{\theta}{2})}|) . When you plug in an imaginary number \frac{i \pi}{\lambda} for p , you actually do get out the correct relationship between an \frac{i \pi}{\lambda} -gon’s edge length and internal angle that can be found from solving the previously considered quadrilaterals. s = 2 {\rm {arccosh}} (|\frac{\cos (\frac{\pi}{\frac{i\pi}{\lambda}})}{\sin (\frac{\theta}{2})}|) \\ s = 2 {\rm {arccosh}} (|\frac{\cos (\frac{\lambda \pi}{i \pi})}{\sin (\frac{\theta}{2})}|) \\ s = 2 {\rm {arccosh}} (|\frac{\cos (\frac{\lambda}{i})}{\sin (\frac{\theta}{2})}|) \\ s = 2 {\rm {arccosh}} (|\frac{\cos (-i \lambda)}{\sin (\frac{\theta}{2})}|) \\ s = 2 {\rm {arccosh}} (|\frac{\cos (i \lambda)}{\sin (\frac{\theta}{2})}|) \\ s = 2 {\rm {arccosh}} (|\frac{\cosh (\lambda)}{\sin (\frac{\theta}{2})}|) A hyperbolic straight line (arcs of circles that cut the boundary circle at right angles in the Poincaré disk) is a special case of hypercycle (arcs of circles that cut the boundary circle), so pseudogons can be circumscribed by them just fine— their edges will all just be collinear and their internal angles will be 180 degrees. Numbers with an infinitely long decimal expansion will be rounded to 5 decimal places. *Name(s); Schläfli symbol(s); Vertex configuration **Derivation t_k \{3, z\} *Bitruncated order-3 4.77098i-gonal tiling, Truncated order-4.77098i triangular tiling; t_{1,2} \{4.77098i, 3\} = t_{0,1} \{3, 4.77098i\} ; (4.77098i).6.6 *Cantellated order-3 4.28677i-gonal tiling; Rhombitri(4.28677i)gonal tiling; t_{0,2} \{4.28677i, 3\} , 3.4.(4.28677i).4 *Cantitruncated order-3 3.26425i-gonal tiling, Truncated tri(3.26425i)gonal tiling; t_{0,1,2} \{3.26425...i, 3\} ; 4.6.(6.5285i) *Snub tri(4.11049i)gonal tiling; sr \{4.11049i, 3\} ; 3.3.3.3.3.(4.11049i) *Tri(2.09513i)trigonal tiling, Cantic 4.19026i-gonal tiling; ; 3.6.(2.09513i).6 *Snub order-5.92041i triangular tiling, Snub tri(2.96021i)trigonal tiling; s\{3, 5.92041i\} ; 3.3.3.3.3.(2.96021i) t_k \{4, z\} *Bitruncated order-4 4.11049i-gonal tiling, Truncated order-4.11049 square tiling; t_{1, 2} \{4.11049i, 4\} = t_{0,1} \{4, 4.11049i\} ; (4.11049i).8.8 *Cantellated order-4 3.5644i-gonal tiling, Rhombitetra(3.5644i)gonal tiling; t_{0,2} \{3.5644i, 4\} ; 4.4.4.(3.5644i) Category:Blog posts